Chapter 7

Div, grad, and curl

7.1

The operator and the gradient:

Recall that the gradient of a differentiable scalar field on an open set D in

Rn is given by the formula:



,
,...,
=
.
(7.1)
x1 x2
xn
It is often convenient to define formally the differential operator in vector
form as:



=
,
,...,
.
(7.2)
x1 x2
xn
Then we may view the gradient of , as the notation suggests, as the
result of multiplying the vector by the scalar field . Note that the order

is not x j .
of multiplication matters, i.e., x
j
Let us now review a couple of facts about the gradient. For any j n,

is identically zero on D iff (x1 , x2 , . . . , xn ) is independent of xj . Consexj

quently,
= 0 on D

= constant.

(7.3)

is normal to the level set Lc ().

(7.4)

Moreover, for any scalar c, we have:

Thus gives the direction of steepest change of .

1

7.2

Divergence

Let f : D Rn , D Rn , be a differentiable vector field. (Note that both

spaces are n-dimensional.) Let f1 , f2 , . . . , fn be the component (scalar) fields
of f . The divergence of f is defined to be
div(f ) = f =

n
X
fj
.
x
j
j=1

(7.5)

This can be reexpressed symbolically in terms of the dot product as

f =(

,...,
) (f1 , . . . , fn ).
x1
xn

(7.6)

Note that div(f ) is a scalar field.

Given any n n matrix A = (aij ), its trace is defined to be:
tr(A) =

n
X

aii .

i=1

Then it is easy to see that, if Df denotes the Jacobian matrix, then

f = tr(Df ).

(7.7)

Let be a twice differentiable scalar field. Then its Laplacian is defined

to be
2 = ().

(7.8)

It follows from (7.1),(7.5),(7.6) that

2 2
2
=
+ 2 + + 2 .
x21
x2
xn
2

(7.9)

One says that is harmonic iff 2 = 0. Note that we can formally

consider the dot product
n

X 2

=(
,...,
)(
,...,
)=
.
x1
xn
x1
xn
x2j
j=1
Then we have
2

(7.10)

2 = ( ).

(7.11)

Examples of harmonic functions:

(i) D = R2 ; (x, y) = ex cos y.
Then
= ex cos y,
= ex sin y,
x
y
2
x2

= ex cos y,

2
y 2

= ex cos y. So, 2 = 0.
p
(ii) D = R2 {0}; (x, y) = log( x2 + y 2 ) = log(r).
(x2 +y 2 )2x(2x)

y
2
x
Then
=
,
=
,
=
=
2
2
2
2
2
x
x +y
y
x +y
x
(x2 +y 2 )2
and

(x2 +y 2 )2y(2y)
(x2 +y 2 )2

(x2 y 2 )
.
(x2 +y 2 )2

(x2 y 2 )
,
(x2 +y 2 )2

and

2
y 2

So, = 0.

These last two examples are special cases of the fact, which we mention
without proof, that for any function f : D C which is differentiable in the
complex sense, the real and imaginary part, <(f ) and =(f ), are harmonic
functions. Here f is differentiable in the complex sense if its total derivative
Df at a point z D, a priori a R-linear map from C to itself, is in fact given
by multiplication with a complex number, which we then call f 0 (z). More
concretely,
this means that the matrix of Df in the basis 1, i is of the form


a b
for some real numbers a, b. We then have f 0 (z) = a + bi. There is
b a
a large supply of such functions since any f given (locally) by a convergent
power series in z is complex differentiable.
In (i) we can take f (z) = ez = ex+iy = ex cos(y) + iex sin(y) and in
(ii) we can take f (z) = log(z) = log(rei ) = log(r) + i but we must be
careful about the domain. To have a well defined argument for all z D
we must make a cut in the plane and can only define f on, for example,
D = {z = x + iy| y = 0 x > 0} or D0 = {z = x + iy| y = 0 x < 0}. But
the union of D and D0 is C {0} as in (ii) .
(iii) D = Rn {0}; (x1 , x2 , ..., xn ) = (x21 + x22 + + x2n )/2 = r for some
fixed R.

= r1 xri = r2 xi , and
Then x
i
2
x2i

= ( 2)r4 xi xi + r2 1.
P
Hence 2 = ni=1 (( 2)r4 x2i + r2 ) = ( 2 + n)r2 .
So is harmonic for = 0 or = 2 n ( = 1 for n = 3).

7.3

Cross product in R3

The three-dimensional space is very special in that it admits a vector product, often called the cross product. Let i,j,k denote the standard basis of
R3 . Then, for all pairs of vectors v = xi + yj + zk and v 0 = x0 i + y 0 j + z 0 k,
the cross product is defined by
v v 0 = det

i j k
x y z
x0 y 0 z 0

= (yz 0 y 0 z)i (xz 0 x0 z)j + (xy 0 x0 y)k. (7.12)

Lemma 1 (a) v v 0 = v 0 v (anti-commutativity) (b) ij = k, jk = i,

k i = j (c) v (v v 0 ) = v 0 (v v 0 ) = 0.
Corollary: v v = 0.
Proof of Lemma (a) v 0 v is obtained by interchanging the second and
third rows of the matrix
determinant gives v v 0 . Thus v 0 v=v v 0 .
 i j kwhose

(b) i j = det 1 0 0 , which is k as asserted. The other two identities
0 1 0
are similar.
(c) v (v v 0 ) = x(yz 0 y 0 z) y(xz 0 x0 z) + z(xy 0 x0 y) = 0. Similarly
for v 0 (v v 0 ).
Geometrically, v v 0 can, thanks to the Lemma, be interpreted as follows.
Consider the plane P in R3 defined by v,v 0 . Then v v 0 will lie along the
normal line to this plane at the origin, and its orientation is given by the right
hand rule: If the fingers of your right hand grab a pole and you view them
from the top as a circle in the v v 0 -plane that is oriented counterclockwise
(i.e. corresponding to the ordering (v, v 0 ) of the basis) then the thumb points
in the direction of v v 0 .
Finally the length ||v v 0 || is equal to the area of the parallelogram
spanned by v and v 0 . Indeed this area is equal to the volume of the parallelepiped spanned by v, v 0 and a unit vector u = (ux , uy , uz ) orthogonal to v
and v 0 . We can take u = v v 0 /||v v 0 || and the (signed) volume equals

ux uy uz
det x y z =ux (yz 0 y 0 z) uy (xz 0 x0 z) + uz (xy 0 x0 y)
x0 y 0 z 0
=||v v 0 || (u2x + u2y + u2z ) = ||v v 0 ||.
4

More generally, the same argument shows that the (signed) volume of the
parallelepiped spanned by any three vectors u, v, v 0 is u (v v 0 ).

Curl of vector fields in R3

7.4

Let f : D R3 , D R3 be a differentiable vector field. Denote by P ,Q,R

its coordinate scalar fields, so that f = P i + Qj + Rk. Then the curl of f is
defined to be:
 i j k

.
(7.13)
curl(f ) = f = det x
y z
P

Q R

Note that it makes sense to denote it f , as it is formally the cross

product of with f .
If the vector field f represents the flow of a fluid, then the curl measures
how the flow rotates the vectors, whence its name.
Proposition 1 Let h (resp. f ) be a C 2 scalar (resp. vector) field. Then
(a) (h) = 0.
(b) ( f ) = 0.
Proof: (a) By definition of gradient and curl,
i

(h) = det

=

2h
2h

yz zy


i+

x y z
h h h
x y z

2h
2h

zx xz


j+

2h
2h

xy yx


k.

Since h is C 2 , its second mixed partial derivatives are independent of the

order in which the partial derivatives are computed. Thus, (f ) = 0.
(b) By the definition of divergence and curl,

 


R Q R P Q P
( f ) =
, ,

,
+
,

x y z
y
z
x
z x
y


=

2R
2Q

xy xz

Again, since f is C 2 ,
Done.


  2

2R
2P
Q
2P
+
+
+

.
yx yz
zx zy

2R
xy

2R
,
yx

etc., and we get the assertion.

Warning: There exist twice differentiable scalar (resp. vector) fields

h (resp. f ), which are not C 2 , for which (a) (resp. (b)) does not hold.
When the vector field f represents fluid flow, it is often called irrotational when its curl is 0. If this flow describes the movement of water in a
stream, for example, to be irrotational means that a small boat being pulled
by the flow will not rotate about its axis. We will see later in this chapter
the condition f = 0 occurs naturally in a purely mathematical setting
as well.
y
Examples: (i) Let D = R3 {0} and f (x, y, z) = (x2 +y
2) i
that f is irrotational. Indeed, by the definition of curl,
!
i
j
k

f = det

=
z

x
j.
(x2 +y 2 )

Show

x
y
z
y
x
0
(x2 +y 2 ) (x2 +y 2 )




 


y
x
y

i+
j+

k
z x2 + y 2
x x2 + y 2
y x2 + y 2


(x2 + y 2 ) + 2x2 (x2 + y 2 ) 2y 2

=
k = 0.
(x2 + y 2 )2
(x2 + y 2 )2

x
2
x + y2

(ii) Let m be any integer 6= 3, D = R3 p

{0}, and
f (x, y, z) = r1m (xi + yj + zk), where r = x2 + y 2 + z 2 . Show that f is not
the curl of another vector field. Indeed, suppose f = g. Then, since f
is C 1 , g will be C 2 , and by the Proposition proved above,
f = ( g) would be zero. But,

 

x y z 
m, m, m
f =
, ,
x y z
r r r
rm 2x2 ( m2 )rm2 rm 2y 2 ( m2 )rm2 rm 2z 2 ( m2 )rm2
=
+
+
r2m
r2m
r2m
6



1
m
2
2
2 m2
= m (3 m).
3r

m(x
+
y
+
z
)r
2m
r
r
This is non-zero as m 6= 3. So f is not a curl.
=

Warning: It may be true that the divergence of f is zero, but f is still

not a curl. In fact this happens in example (ii) above if we allow m = 3. We
cannot treat this case, however, without establishing Stokes theorem.

7.5

An interpretation of Greens theorem via

the curl

Recall that Greens theorem for a plane region with boundary a piecewise
C 1 Jordan curve C says that, given any C 1 vector field g = (P, Q) on an open
set D containing , we have:

I
ZZ 
Q P

dx dy =
P dx + Q dy.
(7.14)
x
y
C

We will now interpret the term Q

P
. To do that, we think of the
x
y
3
:=
plane as sitting in R as {z = 0}, and define a C 1 vector field f on D
3
{(x, y, z) R
|(x, y) 
D} by setting f (x, y, z) = g(x, y) = P i + Qj. Then


i j k
Q
P

k, because P
f = det x
=

= Q
= 0. Thus we
y z
y
x
z
z
P

get:
( f ) k =

Q P

.
y
x

(7.15)

And Greens theorem becomes:

RR
H
Theorem 1 ( f ) k dx dy = C P dx + Q dy

7.6

A criterion for being conservative via the

curl

Here we just reformulate the remark after Ch. 6, Cor. 1 (which we didnt
completely prove but just made plausible) using the curl.
7

Proposition 1 Let g : D R2 , D R2 open and simply connected,

g = (P, Q), be a C 1 vector field. Set f (x, y, z) = g(x, y), for all (x, y, z) R3
with (x, y) D. Suppose f = 0. Then g is conservative on D.
H
Proof: Since f = 0, Theorem 1 implies that C P dx + Q dy = 0 for all
Jordan
curves C contained in D. In fact, f = 0 also implies that
H
P
dx
+ Q dy = 0 for all closed curves but we wont prove this. Hence f is
C
conservative. Done.

Example: D = R2 {(x, 0) R2 | x 0}, g(x, y) =

Determine if g is conservative on D:

y
i
x2 +y 2

x
j.
x2 +y 2

Again, define f (x, y, z) to be g(x, y) for all (x, y, z) in R3 such that (x, y)
D. Since g is evidently C 1 , f will be C 1 as well. By the Proposition above, it
will suffice to check if f is irrotational, i.e., f = 0, on D R. This was
already shown in Example (i) of section 4 of this chapter. So g is conservative.